A particle moves such that its acceleration is given by a=-β (x - 2) Here: β is a positive constant and x is the position from origin. Time period of oscillations is

Question

A particle moves such that its acceleration is given by a=-β (x - 2) Here: β is a positive constant and x is the position from origin. Time period of oscillations is (A) 2π √β  (B) 2π √(1/β ) (C) 2π √β + 2 (D) 2π √(1/β + 2)

Tardigrade Admin · Accepted Answer

a=-β (x - 2) as a=-ω2(x - x0) by comparing ω 2=β ⇒ T=2π √(1/β )

Q. A particle moves such that its acceleration is given by

$a = - β (x - 2)$

Here : $β$ is a positive constant and $x$ is the position from origin. Time period of oscillations is

$2 π β$

$2 π \frac{1}{β}$

$2 π β + 2$

$2 π \frac{1}{β + 2}$

$a = - β (x - 2)$
as $a = - ω^{2} (x - x_{0})$
by comparing
$ω^{2} = β$
$\Rightarrow T = 2 π \frac{1}{β}$

Q. A particle moves such that its acceleration is given by a=−β(x−2) Here : β is a positive constant and x is the position from origin. Time period of oscillations is

2πβ​

2πβ1​​

2πβ+2​

2πβ+21​​